需要对文件中的数组进行模糊匹配

Question

我正在尝试完成家庭作业，我需要将数组中 5 个项目的组合与文件列表匹配。

"Orange, banana, cat, apple, strings"

我有一个包含所有这些单词的文件，以及它们的组合

ex: 
1: Orange
2: Banana
3: CatBanana

我需要单独计算所有这些

`ex: Orange=1, Banana=1 CatBanana=1`

Answer 1

str.count((

使用 str.count(s) 计算 Str 中 s 出现的次数：

datafile = 'data.txt'
items = "Orange banana cat apple strings".lower().split()
counts = dict.fromkeys(items, 0)
with open(datafile, 'rt') as f:
    for line in f:
        line = line.lower()
        for item in items:
            counts[item] += line.count(item)
print(counts)

Answer 2

我假设您正在尝试查找单词列表在文件中出现的次数(逐行(？

w = ['some', 'list', 'of', 'words']
r = dict.fromkeys(w,0)  # used to store values
filename = 'file.txt'
with open(filename, 'r') as f:  # open file
    lines = f.readlines()  # file to list
    for string in lines:
        for word in w:
            if word in string:
                r[word] += 1  # update
print(r)

如果没有，那么我会看看这篇文章：将文件中的单词读入字典

编辑：

只需格式化单词

数组以匹配您尝试查找的单词或小写所有内容，这样格式化就不会妨碍(就像 RootTwo 所做的那样(。我相信您也在尝试将文件与数组进行比较，以便：

# fix formatting for comparison.
format_this = "Orange, banana, cat, apple, strings"  # lowercase if necessary
separate = format_this.split(', ')  # ['Orange', 'banana', 'cat', 'apple', 'strings']
compare = [word.capitalize() for word in separate]  # ['Orange', 'Banana', 'Cat', 'Apple', 'Strings']
# set file to dict.
filename = 'file.txt'
lines = []
with open(filename, 'r') as f:  # open file
    lines = f.readlines()
lines = [line.strip('n') for line in lines]  # clean up, lowercase if necessary
r = dict.fromkeys(lines,0)  # {'Orange': 0, 'Banana': 0, 'CatBanana': 0}
# compare and record.
for key in r:
    for word in compare:
        if word in key:  # match found
            r[key] += 1  # update value
            break  # stop and move on once match found, remove if necessary
print(r)  # {'Orange': 1, 'Banana': 1, 'CatBanana': 1}